Subj : Re: Memory Pointers and Free
To   : borland.public.cpp.borlandcpp
From : waking
Date : Sun Jul 25 2004 09:18 am

On Sun, 25 Jul 2004 01:27:49 +0100, =?ISO-8859-15?Q?Jo=E3o?= Paredes
<joaommp@myrealbox.com> wrote:

>> s_a *a = malloc(sizeof(s_a));
>> a->somedata = malloc(512);

>with the declaration for the structure and variable you posted, I don't
>believe that would even compile. 

<chuckle> Interesting example of the differences between C and C++.
You're right, it wouldn't compile. But for different reasons in the two
languages.

>If you create a type with "struct something
>{...};" you would have to declare variables with "struct something myvar;". 

True for C (C90), but not for C++. In the former, this is an error:
 s_a *a = malloc(sizeof(s_a));
It must be this for C90:
struct s_a *a = malloc(sizeof(struct s_a));

However, in C++ the original post is OK as the "struct" keyword
may be omitted. However, in C++ the original example will fail
for another reason - which doesn't apply to C90 - the void*
returned by malloc must be typecast in C++ but may be
omitted in C90. So C++ would look like this:
s_a *a = (s_a*)malloc(sizeof(s_a));

Note that C++ will also accept the explicit "struct" keyword,
and C will accept the typecast. So a form which would pass
both C and C++ would be like this:
struct s_a *a = (struct s_a*)malloc(sizeof(struct s_a));


--
Wayne A. King 
(waking@idirect.com, Wayne_A_King@compuserve.com)

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